1.

Which of the following statements are correct ? (I)Both melting and boiling points of H_2O are higher than those of H_2Te. (II)In both N_2O_5 and N_2O_4 all N-O bond lengths are equivalent (III)In both crystalline NaHCO_3, HCO_3^(-) forms only dimeric anion through hydrogen bond. (IV)Amongst B_2, C_2, N_2^(-) and O_2 on further ionization (losing single electron) form thermodynamically more stable species.

Answer»

(I) and (II)
(III) and (IV)
(II) and (III)
(I) and (IV)

Solution :(I)On account of H-bonding `H_2O` has higher melting and boiling points.
`{:(,H_2O,H_2Te),("m.p/K",273,222),("b.p/K",373,269):}`
(II)
(III)(a) In `NaHCO_3` , the `HCO_3^-` ions are linked into an infinite CHAIN and (b) in `KHCO_3, HCO_3^-` FORMS a dimeric anion.

(IV)`B_2-B.O.=1 and B_2^(+) - B.O=1/2 , C_2-B.O. =2 and C_2^(+) B.O.=1.5`
`N_2^(-)-B.O=2.5 and N_2-B.O =3 , O_2-B.O=2 and O_2^(+)B.O.=2.5`
Bond strength `prop` bond ORDER.


Discussion

No Comment Found